HDU_1312——二维空间BFS
这题网上都是用DFS做的,我也不知道怎么用BFS做的。。。有个注意点,就是一步也走不动的时候,输出1
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #define MAX 21 5 using namespace std; 6 struct point 7 { 8 int x,y; 9 }start; 10 11 int N,M,dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; 12 char map[MAX][MAX]; 13 14 int Bfs(void); 15 int main() 16 { 17 while(~scanf("%d %d",&M,&N) && N!=0&&M!=0) 18 { 19 memset(map,0,sizeof(map)); 20 for(int i=0;i<N;i++) 21 { 22 scanf("%s",map[i]); 23 for(int j=0;j<M;j++) 24 { 25 if(map[i][j]=='@') 26 start.x=i,start.y=j; 27 } 28 } 29 int num=Bfs(); 30 printf(num==0?"1\n":"%d\n",num); 31 } 32 return 0; 33 } 34 35 int Bfs(void) 36 { 37 int move=0; 38 queue<point>p; 39 p.push(start); 40 point temp,next; 41 while(!p.empty()) 42 { 43 temp=p.front(); 44 p.pop(); 45 for(int i=0,flag=0;i<4;i++) 46 { 47 next.x=temp.x+dir[i][0]; 48 next.y=temp.y+dir[i][1]; 49 if(next.x>=0&&next.x<N && next.y>=0&&next.y<M && map[next.x][next.y]!='#') 50 { 51 map[next.x][next.y]='#'; 52 move++; 53 p.push(next); 54 } 55 } 56 } 57 return move; 58 }
——现在的努力是为了小时候吹过的牛B!!